Finding min/max/saddle $z=x^2-y^2$ Question:

Consider $z=x^2-y^2$. 

$\frac{\partial z}{\partial x}=2x \Rightarrow 2x=0 \Rightarrow x=0$
$\frac{\partial z}{\partial y}=-2y\Rightarrow -2y=0 \Rightarrow y=0$
$\frac{\partial^2 z}{\partial x^2}=2>0$
$\frac{\partial^2 z}{\partial y^2}=-2<0$
Can I stop here and say that $(0,0,0)$ is a saddle point?
 A: Not quite. 
Writing $f(x,y) = x^2 - y^2$ we have $\frac{\partial^2 f(x,y)}{\partial x \partial y} = \frac{\partial^2 f(x,y)}{\partial y \partial x} = 0 $ and as you have found $\frac{\partial^2 f(x,y)}{\partial x^2}= 2$ and $\frac{\partial^2 f(x,y)}{\partial y^2}= -2$, then 
1) The Hessian matrix at $(0,0)$ is given by $$Hf(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix}$$
Thus its eigenvalues are $\lambda_1 = 2> 0$ and $\lambda_2 = -2 < 0$, then  it is a saddle point. 
2) You could also notice that for any disk centered at $0$ the function $f$ assume values greater or less than $0 = f(0,0)$, then it wouldn't be neither a maximum nor a minimum. 
A: no you must form Hessian and say Hessian is not positive or negative definite hence it's a saddle point.
$H = \begin{pmatrix}
        \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial x\partial y}\\
        \frac{\partial^2 z}{\partial y\partial x} & \frac{\partial^2 z}{\partial y^2}\\
        \end{pmatrix}$
$$det(H)<0$$ so eigenvalues are not co-sign. 
A: Adding that the mixed second order derivatives vanish, you have found that the function is stationary at the origin with Hessian $\DeclareMathOperator{diag}{diag}H=\diag(2,-2)$. As $e_1^t H e_1 = 2 > 0$ and $e_2^t H e_2 = -2 < 0$ $H$ is indefinite, and you found a saddle point there.

A: Normally, you will have to calculate the Hessian matrix $H=\begin{bmatrix}\frac{\partial^2z}{\partial x^2} &\frac{\partial^2z}{\partial x\partial y}\\ \frac{\partial^2z}{\partial x\partial y} &\frac{\partial^2z}{\partial x^2}\end{bmatrix}$ and determine if $H$ evaluated in the critical point is positive definite, negative definite, or neither of those (in which case you will say that the critical point is a saddle point).
However, if $z=f(x,y)$ is a function of class $C^2$ (the second order derivatives exists and are continuous) then you will have that $$\dfrac{\partial^2z}{\partial x\partial y}=\dfrac{\partial^2z}{\partial y\partial x}$$ and $$\det(H)=\dfrac{\partial^2z}{\partial x^2}\cdot\dfrac{\partial^2z}{\partial y^2}-\left(\dfrac{\partial^2z}{\partial x\partial y}\right)^2$$
Therefore, when you have a function that is $C^2$ (like $z=x^2-y^2$, or like most of the functions in a course of multivariate calculus) we can conclude the following:
If you already know that $\dfrac{\partial^2z}{\partial x^2},\dfrac{\partial^2z}{\partial y^2}$ have different signs, then the product $\dfrac{\partial^2z}{\partial x^2}\cdot\dfrac{\partial^2z}{\partial y^2}$ is negative and $\det(H)$ will be negative independently of the value of $\dfrac{\partial^2z}{\partial x\partial y}$. Thus, the Hessian matrix $H$ will be neither positive nor negative definite.
So, to conclude, my answer would be that you can stop there and conclude that your point is a saddle point, but only if you really understand why.
